图像处理|Opencv学习笔记透视变换(perspective transform) opencv|透视变换|图像处理|Open

拉伸、收缩、扭曲、旋转是图像的几何变换，在三维视觉技术中大量应用到这些变换，又分为仿射变换和透视变换。仿射变换通常用单应性建模，利用cvWarpAffine解决密集映射，用cvTransform解决稀疏映射。仿射变换可以将矩形转换成平行四边形，它可以将矩形的边压扁但必须保持边是平行的，也可以将矩形旋转或者按比例变化。透视变换提供了更大的灵活性，一个透视变换可以将矩阵转变成梯形。当然，平行四边形也是梯形，所以仿射变换是透视变换的子集。
opencv中的函数主要是：
对图像进行透视变换
void cvWarpPerspective( const CvArr* src, CvArr* dst,const CvMat* map_matrix,
int flags=CV_INTER_LINEAR+CV_WARP_FILL_OUTLIERS, CvScalar fillval=cvScalarAll(0) );
【图像处理|Opencv学习笔记透视变换(perspective transform)】由四对点计算透射变换
CvMat* cvGetPerspectiveTransform( const CvPoint2D32f*src, const CvPoint2D32f* dst, CvMat*map_matrix );
透视变换(Perspective Transformation)是将成像投影到一个新的视平面(Viewing Plane)，也称作投影映射(Projective Mapping)。如图1，通过透视变换ABC变换到A'B'C'。

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参考代码如下：

# 透视变换 # import the necessary packages import numpy as np import argparse import cv2 def order_points(pts): # initialzie a list of coordinates that will be ordered # such that the first entry in the list is the top-left, # the second entry is the top-right, the third is the # bottom-right, and the fourth is the bottom-left rect = np.zeros((4, 2), dtype = "float32") # the top-left point will have the smallest sum, whereas # the bottom-right point will have the largest sum s = pts.sum(axis = 1) rect[0] = pts[np.argmin(s)] rect[2] = pts[np.argmax(s)] # now, compute the difference between the points, the # top-right point will have the smallest difference, # whereas the bottom-left will have the largest difference diff = np.diff(pts, axis = 1) rect[1] = pts[np.argmin(diff)] rect[3] = pts[np.argmax(diff)] # return the ordered coordinates return rectdef four_point_transform(image, pts): # obtain a consistent order of the points and unpack them # individually rect = order_points(pts) (tl, tr, br, bl) = rect # compute the width of the new image, which will be the # maximum distance between bottom-right and bottom-left # x-coordiates or the top-right and top-left x-coordinates widthA = np.sqrt(((br[0] - bl[0]) ** 2) + ((br[1] - bl[1]) ** 2)) widthB = np.sqrt(((tr[0] - tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2)) maxWidth = max(int(widthA), int(widthB)) # compute the height of the new image, which will be the # maximum distance between the top-right and bottom-right # y-coordinates or the top-left and bottom-left y-coordinates heightA = np.sqrt(((tr[0] - br[0]) ** 2) + ((tr[1] - br[1]) ** 2)) heightB = np.sqrt(((tl[0] - bl[0]) ** 2) + ((tl[1] - bl[1]) ** 2)) maxHeight = max(int(heightA), int(heightB)) # now that we have the dimensions of the new image, construct # the set of destination points to obtain a "birds eye view", # (i.e. top-down view) of the image, again specifying points # in the top-left, top-right, bottom-right, and bottom-left # order dst = np.array([ [0, 0], [maxWidth - 1, 0], [maxWidth - 1, maxHeight - 1], [0, maxHeight - 1]], dtype = "float32") # compute the perspective transform matrix and then apply it M = cv2.getPerspectiveTransform(rect, dst) warped = cv2.warpPerspective(image, M, (maxWidth, maxHeight)) # return the warped image return warped# load the image and grab the source coordinates (i.e. the list of # of (x, y) points) # NOTE: using the 'eval' function is bad form, but for this example # let's just roll with it -- in future posts I'll show you how to # automatically determine the coordinates without pre-supplying them image = cv2.imread("C:/Users/zyh/Desktop/20201004114816.png") pts = np.array([[73, 239], [356, 117], [475, 265], [187, 443]], dtype = "float32") # apply the four point tranform to obtain a "birds eye view" of # the image warped = four_point_transform(image, pts) # show the original and warped images cv2.imshow("Original", image) cv2.imshow("Warped", warped) cv2.waitKey(0)

图像处理|Opencv学习笔记透视变换(perspective transform)

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原图

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变换后的图
参考如下：
perspective transform（透视变换）的实现过程_一只安静的大白的博客-CSDN博客_perspectivetransform
【OpenCV3】透视变换——cv::getPerspectiveTransform()与cv::warpPerspective()详解 - 程序员大本营